Class 12 FSc Math Chapter 4 Introduction to Analytic Geometry Straight Lines Solved Questions PDF Guide | Punjab Boards & FBISE

Theoretical Foundations & Core Theorems

Analytic Geometry (or Coordinate Geometry) bridges algebra and geometry by utilizing a rectangular Cartesian coordinate system. In two-dimensional space $\mathbb{R}^2$, every point $P$ is represented uniquely by an ordered pair of real numbers $(x, y)$, and every geometric locus corresponds to an algebraic equation.

1. Fundamental Distance & Ratio Principles

For any two points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the Cartesian plane, the distance $d$ between them is derived directly from the Pythagorean theorem:

$$d = |AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

The Ratio Formula (Section Formula): If a point $P(x, y)$ divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $k_1 : k_2$, its coordinates are given by:

$$x = \frac{k_1 x_2 + k_2 x_1}{k_1 + k_2}, \quad y = \frac{k_1 y_2 + k_2 y_1}{k_1 + k_2}$$

• Internal Division: Ratio $k_1 : k_2 > 0$ (the point $P$ lies between $A$ and $B$).
• External Division: Ratio is negative, i.e., $k_1 : -k_2$ (the point $P$ lies on the extension of line segment $AB$).
• Midpoint Formula: When $k_1 : k_2 = 1 : 1$, $P = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.

2. Inclination and Slope (Gradient) of a Line

The inclination $\alpha$ of a non-horizontal line is the angle measured counterclockwise from the positive $x$-axis to the line ($0^\circ \le \alpha < 180^\circ$). The slope $m$ of the line is defined as:

$$m = \tan \alpha = \frac{y_2 - y_1}{x_2 - x_1} \quad (x_1 \neq x_2)$$

• If two lines with slopes $m_1$ and $m_2$ are parallel ($L_1 \parallel L_2$), then $m_1 = m_2$.
• If two non-vertical lines are perpendicular ($L_1 \perp L_2$), then $m_1 \cdot m_2 = -1 \iff m_2 = -\frac{1}{m_1}$.

3. Analytical Equations of Straight Lines

Depending on the given geometric parameters, a straight line can be expressed in several standard algebraic forms:

1. Slope-Intercept Form: $y = mx + c$, where $m$ is the slope and $c$ is the $y$-intercept.
2. Point-Slope Form: $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope.
3. Two-Point Form: $y - y_1 = \left(\frac{y_2 - y_1}{x_2 - x_1}\right)(x - x_1)$.
4. Two-Intercept Form: $\frac{x}{a} + \frac{y}{b} = 1$, where $a \neq 0$ and $b \neq 0$ are the $x$-intercept and $y$-intercept respectively.
5. Normal (Symmetric) Form: $x \cos \alpha + y \sin \alpha = p$, where $p > 0$ is the length of the perpendicular segment from the origin to the line, and $\alpha$ is the angle this perpendicular makes with the positive $x$-axis.
6. General Linear Equation: $Ax + By + C = 0$ (where $A$ and $B$ are not both zero). Its slope is $m = -\frac{A}{B}$, its $x$-intercept is $-\frac{C}{A}$, and its $y$-intercept is $-\frac{C}{B}$.

4. Perpendicular Distance and Angle Between Two Lines

The absolute perpendicular distance $d$ from a given point $P(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by:

$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

The acute angle $\theta$ between two non-vertical intersecting lines $L_1$ and $L_2$ with slopes $m_1$ and $m_2$ is determined by:

$$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \quad (m_1 m_2 \neq -1)$$

5. Homogeneous Second-Degree Equations in Two Variables

An equation of the form $ax^2 + 2hxy + by^2 = 0$ represents a pair of straight lines passing through the origin $(0,0)$.

• The individual slopes $m_1, m_2$ of these two lines satisfy $m_1 + m_2 = -\frac{2h}{b}$ and $m_1 m_2 = \frac{a}{b}$.
• The angle $\theta$ between the two lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by:

$$\tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b}$$

• The lines are real and distinct if $h^2 - ab > 0$.
• The lines are coincident if $h^2 - ab = 0$.
• The lines are imaginary (except for the origin) if $h^2 - ab < 0$.
• The lines are perpendicular if $a + b = 0$.

Formula Summary & Quick Reference

Concept / Topic	Standard Formula / Condition	Primary Application / Remarks
Distance Formula	$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$	Calculating spatial distance between two points in $\mathbb{R}^2$.
Internal Ratio Formula	$\left(\frac{k_1 x_2 + k_2 x_1}{k_1 + k_2}, \frac{k_1 y_2 + k_2 y_1}{k_1 + k_2}\right)$	Point dividing segment $AB$ internally in ratio $k_1 : k_2$.
External Ratio Formula	$\left(\frac{k_1 x_2 - k_2 x_1}{k_1 - k_2}, \frac{k_1 y_2 - k_2 y_1}{k_1 - k_2}\right)$	Point dividing segment $AB$ externally in ratio $k_1 : k_2$.
Centroid of Triangle	$\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)$	Intersection point of the three medians of a triangle.
Incenter of Triangle	$\left(\frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c}\right)$	Intersection of internal angle bisectors ($a, b, c$ are side lengths).
Slope ($m$)	$m = \tan \alpha = \frac{y_2 - y_1}{x_2 - x_1} = -\frac{A}{B}$	Measures rate of change / inclination of a straight line.
Perpendicular Condition	$m_1 \cdot m_2 = -1 \iff A_1 A_2 + B_1 B_2 = 0$	Determining orthogonal lines.
Parallel Condition	$m_1 = m_2 \iff \frac{A_1}{A_2} = \frac{B_1}{B_2}$	Determining parallel lines.
Normal Form	$x \cos \alpha + y \sin \alpha = p$	Line specification via length $p$ of perpendicular from origin.
Point-to-Line Distance	$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$	Shortest distance from point $(x_1, y_1)$ to line $Ax+By+C=0$.
Area of Triangle	$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$	Calculates enclosed area; area $= 0$ proves collinearity.
Angle Between Lines	$\tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|$	Finds acute angle between two intersecting lines.
Pair of Lines Angle	$\tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b}$	Angle between lines given by $ax^2 + 2hxy + by^2 = 0$.

Step-by-Step Solved Board Exam Questions

Question 1 (Section Formula & Geometric Applications)

Problem Statement: Find the coordinates of the point $P$ that divides the segment joining $A(-6, 3)$ and $B(5, -2)$ internally in the ratio $2:3$, and the point $Q$ that divides it externally in the ratio $2:3$. Furthermore, find the centroid of triangle $ABC$ where $C(4, 8)$.

Solution:

Part A: Internal Division Point $P(x, y)$

Given coordinates: $A(x_1, y_1) = (-6, 3)$, $B(x_2, y_2) = (5, -2)$, and ratio $k_1 = 2, k_2 = 3$.

Using the internal ratio formula:

$$x = \frac{k_1 x_2 + k_2 x_1}{k_1 + k_2} = \frac{2(5) + 3(-6)}{2 + 3} = \frac{10 - 18}{5} = -\frac{8}{5}$$ $$y = \frac{k_1 y_2 + k_2 y_1}{k_1 + k_2} = \frac{2(-2) + 3(3)}{2 + 3} = \frac{-4 + 9}{5} = \frac{5}{5} = 1$$

Thus, $P = \left(-\frac{8}{5}, 1\right)$.

Part B: External Division Point $Q(x', y')$

Using the external ratio formula (replacing $k_2$ with $-k_2$):

$$x' = \frac{k_1 x_2 - k_2 x_1}{k_1 - k_2} = \frac{2(5) - 3(-6)}{2 - 3} = \frac{10 + 18}{-1} = -28$$ $$y' = \frac{k_1 y_2 - k_2 y_1}{k_1 - k_2} = \frac{2(-2) - 3(3)}{2 - 3} = \frac{-4 - 9}{-1} = 13$$

Thus, $Q = (-28, 13)$.

Part C: Centroid $G(x_g, y_g)$ of $\Delta ABC$

Vertices: $A(-6, 3)$, $B(5, -2)$, $C(4, 8)$.

$$x_g = \frac{x_1 + x_2 + x_3}{3} = \frac{-6 + 5 + 4}{3} = \frac{3}{3} = 1$$ $$y_g = \frac{y_1 + y_2 + y_3}{3} = \frac{3 + (-2) + 8}{3} = \frac{9}{3} = 3$$

Thus, the centroid is $G(1, 3)$.

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Question 2 (Intersection of Lines and Perpendicular Condition)

Problem Statement: Find the equation of the line passing through the point of intersection of lines $L_1: 2x - 3y + 4 = 0$ and $L_2: 3x + 4y - 5 = 0$, and perpendicular to the line $L_3: 6x - 7y + 8 = 0$.

Solution:

Method (Family of Lines Technique):

The equation of any line passing through the intersection of $L_1$ and $L_2$ is given by $L_1 + k L_2 = 0$ where $k$ is a scalar parameter:

$$(2x - 3y + 4) + k(3x + 4y - 5) = 0$$

Rearranging in general standard linear form $(Ax + By + C = 0)$:

$$(2 + 3k)x + (-3 + 4k)y + (4 - 5k) = 0 \quad \text{--- (Equation 1)}$$

The slope $m$ of this family of lines is:

$$m = -\frac{A}{B} = -\frac{2 + 3k}{-3 + 4k} = \frac{2 + 3k}{3 - 4k}$$

Now, consider the given line $L_3: 6x - 7y + 8 = 0$. Its slope $m_3$ is:

$$m_3 = -\frac{6}{-7} = \frac{6}{7}$$

Since the required line is perpendicular to $L_3$, their slopes must satisfy $m \cdot m_3 = -1$:

$$\left(\frac{2 + 3k}{3 - 4k}\right) \cdot \left(\frac{6}{7}\right) = -1$$

Solving for $k$:

$$6(2 + 3k) = -7(3 - 4k)$$ $$12 + 18k = -21 + 28k$$ $$12 + 21 = 28k - 18k$$ $$33 = 10k \implies k = \frac{33}{10}$$

Substitute $k = \frac{33}{10}$ back into Equation 1:

$$(2x - 3y + 4) + \frac{33}{10}(3x + 4y - 5) = 0$$ $$10(2x - 3y + 4) + 33(3x + 4y - 5) = 0$$ $$(20x - 30y + 40) + (99x + 132y - 165) = 0$$ $$119x + 102y - 125 = 0$$

Final Answer: The required equation of the line is $119x + 102y - 125 = 0$.

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Question 3 (Perpendicular Distance and Area of Triangle)

Problem Statement: Find the perpendicular distance from the point $P(2, 3)$ to the line $12x - 5y + 6 = 0$. Furthermore, calculate the area of the triangular region with vertices $A(1, 4)$, $B(2, -3)$, and $C(-1, -2)$.

Solution:

Part A: Perpendicular Distance $d$

Given line: $Ax + By + C = 0 \implies 12x - 5y + 6 = 0$, where $A = 12, B = -5, C = 6$.

Given point: $P(x_1, y_1) = (2, 3)$.

Using the point-to-line distance formula:

$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} = \frac{|12(2) + (-5)(3) + 6|}{\sqrt{12^2 + (-5)^2}}$$ $$d = \frac{|24 - 15 + 6|}{\sqrt{144 + 25}} = \frac{|15|}{\sqrt{169}} = \frac{15}{13} \text{ units}$$

Part B: Area of Triangle $ABC$

Vertices: $A(1, 4)$, $B(2, -3)$, $C(-1, -2)$.

The area formula using determinants is:

$$\text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|$$ $$\text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} 1 & 4 & 1 \\ 2 & -3 & 1 \\ -1 & -2 & 1 \end{pmatrix} \right|$$

Expanding the determinant along Row 1:

$$\det = 1[(-3)(1) - (1)(-2)] - 4[(2)(1) - (1)(-1)] + 1[(2)(-2) - (-3)(-1)]$$ $$\det = 1[-3 + 2] - 4[2 + 1] + 1[-4 - 3]$$ $$\det = 1(-1) - 4(3) + 1(-7) = -1 - 12 - 7 = -20$$

Taking the absolute value:

$$\text{Area} = \frac{1}{2} |-20| = \frac{20}{2} = 10 \text{ square units}$$ ---

Question 4 (Homogeneous 2nd Degree Equations & Pair of Straight Lines)

Problem Statement: Find the component lines represented by the homogeneous second-degree equation $2x^2 - 5xy + 3y^2 = 0$, and evaluate the acute angle $\theta$ between them.

Solution:

Part A: Finding the Individual Straight Lines

Given: $2x^2 - 5xy + 3y^2 = 0$. Factorize by splitting the middle term:

$$2x^2 - 2xy - 3xy + 3y^2 = 0$$ $$2x(x - y) - 3y(x - y) = 0$$ $$(2x - 3y)(x - y) = 0$$

Setting each linear factor equal to zero yields the two lines:

$$\text{Line 1: } x - y = 0 \implies y = x \quad (\text{Slope } m_1 = 1)$$ $$\text{Line 2: } 2x - 3y = 0 \implies y = \frac{2}{3}x \quad \left(\text{Slope } m_2 = \frac{2}{3}\right)$$

Part B: Calculating the Angle $\theta$ Between the Lines

Comparing $2x^2 - 5xy + 3y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$ gives:

$$a = 2, \quad 2h = -5 \implies h = -\frac{5}{2}, \quad b = 3$$

Use the angle formula for homogeneous second-degree equations:

$$\tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b}$$

Substitute values:

$$h^2 - ab = \left(-\frac{5}{2}\right)^2 - (2)(3) = \frac{25}{4} - 6 = \frac{25 - 24}{4} = \frac{1}{4}$$ $$\sqrt{h^2 - ab} = \sqrt{\frac{1}{4}} = \frac{1}{2}$$ $$a + b = 2 + 3 = 5$$

Therefore:

$$\tan \theta = \frac{2 \left(\frac{1}{2}\right)}{5} = \frac{1}{5}$$ $$\theta = \tan^{-1}\left(\frac{1}{5}\right) \approx 11.31^\circ$$

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

1. How do I distinguish between internal and external division in ratio problems?

When using the section formula $x = \frac{k_1 x_2 + k_2 x_1}{k_1 + k_2}$, an internal division means the point lies strictly on the line segment connecting $A$ and $B$, resulting in positive scalar values for $k_1$ and $k_2$. An external division means the point lies outside the segment $AB$ on its extended line. Mathematically, external division is handled either by replacing $k_2$ with $-k_2$ in the formula or interpreting a negative calculated ratio $k_1/k_2 < 0$ as external division.

2. How should I correctly handle signs when converting $Ax + By + C = 0$ into Normal Form ($x \cos \alpha + y \sin \alpha = p$)?

To reduce $Ax + By + C = 0$ to Normal Form, first transpose the constant term $C$ to the right side: $Ax + By = -C$. Because the perpendicular distance $p$ from the origin must always be positive ($p > 0$), the right-hand side constant must be positive. If $-C$ is negative, multiply the entire equation by $-1$. Then divide every term by $\sqrt{A^2 + B^2}$. The sign of the denominator $\sqrt{A^2 + B^2}$ is chosen opposite to the sign of $C$.

3. Do board examiners award full marks if I solve "intersection line" problems using point calculation instead of the line family method ($L_1 + k L_2 = 0$)?

Yes, solving for the explicit intersection point $(x_0, y_0)$ by simultaneous linear equations and then finding line equations using point-slope form is mathematically valid and yields full credit in BISE/FBISE board examinations. However, utilizing the family of lines technique $L_1 + k L_2 = 0$ is computationally faster, less prone to arithmetic fraction errors, and explicitly recommended in FSc textbooks for high-scoring direct derivations.